format-specifiers

In an example of "C Primer Plus", the author has used %ul format specifier (in both scanf and printf) for unsigned long . When I try to generalize the problem, it seems that the %ul makes something wrong in my computer. But using %lu solved the issue. Actually, rather than focusing on the problem and the line of codes, I want to know about the difference between %ul and %lu . Maybe I could figure out what's wrong. Searching doesn't give me something useful (except that "they are different"). Any explanation or link/reference is appreciated. %lu is correct, while %ul is incorrect . A printf

I am getting output 0.23 from second printf . But typecasting gives required output. If I am not using type casting previous value is printed. Compiler version is GCC 6.3 #include <stdio.h> int main() { printf("%f ", 0.23); printf("%f", 0); return 0; } LINK FOR IDE in > printf("%f",0); You ask to print a double but you give an int , this is contradictory You are not in the case where the generated code makes a double from the int because printf is not int printf(const char *, double); but int printf ( const char * format, ... ); and the compiler does not look at the format to make the

#include <stdio.h> int main() { int i=10,j=20; printf("%d%d%d",i,j); printf("%d",i,j); return 0; } Using the Turbo C compiler, the output is like: 10 10 garbageValue 20 Can someone explain why this is? Srinath Reddy Dudi Which compiler you are using? I tested it on turbo c v2.0 and turbo c++ v 4.5 compilers and the output is 10 20 garbage value 10 This is the actual output you will get as you are using only two variables and three format specifiers. So it will print the values stored in the two variables and print a garbage value. In the second case you are using only one format specifier and

This question already has an answer here: Using %f to print an integer variable 6 answers May be this is a simple question but I am not sure about how float variables are stored in memory and why it is behaving in this way, can someone please explain about the following behavior. #include<stdio.h> int main () { int a = 9/5; printf("%f\n", a); return 0; } Output: 0.000000 I have looked at some information on how float variables are stored in memory, it has stuff about mantissa, exponent and sign. But I am not getting how to relate that here. int a = 9/5; performs integer division and ignores

Is there any way to have a user inputed float format specifier? For example, if I print this. float c = 15.0123 printf("%.2f", c); // outputs: 15.01 How can I assign the number of decimal places to a variable? Like: int n = 3; float c = 15.0123 printf("%.(%i)f", n, c); // outputs: 15.012 The precision can be specified by an argument with the asterisk * . This is called an argument-supplied precision. float c = 15.0123; int m = 2; printf("%.*f", m, c); printf("%.*f", n, c); that will print out c with n places after the decimal. 来源： https://stackoverflow.com/questions/9627075/dynamic-float

I know that when using %x with printf() we are printing 4 bytes (an int in hexadecimal) from the stack. But I would like to print only 1 byte. Is there a way to do this ? Assumption:You want to print the value of a variable of 1 byte width, i.e., char . In case you have a char variable say, char x = 0; and want to print the value, use %hhx format specifier with printf() . Something like printf("%hhx", x); Otherwise, due to default argument promotion, a statement like printf("%x", x); would also be correct, as printf() will not read the sizeof(unsigned int) from stack , the value of x will be

I read here that there is a sequence point: After the action associated with input/output conversion format specifier. For example, in the expression printf("foo %n %d", &a, 42) , there is a sequence point after the %n is evaluated before printing 42 . However, when I run this code : int your_function(int a, int b) { return a - b; } int main(void) { int i = 10; printf("%d - %d - %d\n", i, your_function(++i, ++i), i); } Instead of what I expect I get: 12 - 0 - 12 Meaning that there was not a sequence point created for the conversion format specifier. Is http://en.wikipedia.org wrong, or have I